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Ooh, I love Professor Layton LPs! Bookmarking this. EDIT: Let's solve some puzzles. 049: B. The message is 'L-A-Y-T-O-N'. 050: The hospital, of course! Very funny, game. 051: (I'll list only the turns) Left, Right, Left, Right, Right, Left, Left, Left, Left, Left, Left, Right. 052: B. Don't ask me to explain this one. Quackles fucked around with this message at 08:58 on Aug 22, 2019 |
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Aug 22, 2019 08:46
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| # ? May 3, 2024 22:50 |
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Thanks for your patience. We should now be returning to our regularly scheduled (?) program.
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Aug 26, 2019 01:09
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wait... is that actually Stachenscarfen's hair? I always though it was suppose to be something like a fur cap.
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Aug 26, 2019 04:16
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053: Third row, fourth through sixth columns and fourth row, sixth and seventh columns as a single piece should slot into the opening in the middle and make the thing a square. 134: I would guess one of the hangers, but I'm not sure which one in particular. 054: Going from the top: ham - lettuce - tomato - potato salad - cheese. 055: Top row: A D E Bottom row: B C F
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Aug 26, 2019 04:17
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Thank you for another scintillating update! For Puzzle 053:  Red indicates where to cut, black indicates where to place. For 134: B. I see a question mark in that hanger! For 054: Okay, let's start with the "the only thing between the ham and tomato is lettuce". Keeping in mind that the potato salad goes after the ham, and the cheese is by the bread, we have the following configuration: Cheese? (Ham-Lettuce-Tomato or Tomato-Lettuce-Ham) PotatoSalad Cheese? If the cheese is on the top then the middle will always be lettuce, which is impossible. So the cheese goes on the bottom. (Ham-Lettuce-Tomato or Tomato-Lettuce-Ham)-PotatoSalad-Cheese Ham's not in the center either, so Tomato-Lettuce-Ham is impossible. Thus it's Ham-Lettuce-Tomato-PotatoSalad-Cheese. For 055: If you map out where each person MIGHT be OK with sitting, you'll notice that none of C, D, E, or F will sit on the left side. So, A and B must be on the left side. A is in front, and B is in the back. Next, C must be in the center column, but not in the front row - they'd be next to A, who they hate. So, C must be in the center back. F will only sit in back center or right, so F has to sit in back right. E wants to sit on the right side, and since F has taken one of the seats in that column, they end up with front right. D ends up front and center. Final arrangement: ADE BCF
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Aug 26, 2019 06:27
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053:  134: Um... what? Even after reading both clues I have no idea what they're going for here. 054: Middle is not ham, and ham above potato salad, so bottom is not ham either. If ham was second from bottom, then lettuce would have to be middle, which violates clue 4. Likewise if ham was second from top, then lettuce would have to be middle. So the ham is on top, and the rest of the ordering comes out easily. From top to bottom: ham, lettuce, tomato, potato salad, cheese. 055: People who don't like sitting on the left: C, D, F, and E. So we have to put A and B on the left: B in back left, and A in front left. Now C wants the middle but can't be in front middle or A will be unhappy, so we put C in back middle. Now F doesn't like the front, so the only place left to put F is back right. That leaves front middle and front right to divide between D and E, and E wants front right so D gets front middle. ADE BCF
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Aug 26, 2019 12:30
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53: Well, given we need to fill that hole, the section we cut off from row 3 needs to be 3 spaces wide. That means that we need a line 2 spaces wide from the row below it. 134: Well, if it's not one of the clothes, I guess the only thing left is the hangars? I guess that means it's C, for the question mark (sorta) that's there? 54: Oh, thank god it's back to these. So we know from clue 2 that it has to be either ham > lettuce > tomato, or the reverse. However, we know from the last clue that the middle one can't be ham or lettuce. So it's tomato in the middle slot. From there, we know from the first clue that ham has to be above potato salad, so ham must be the top layer. That leaves cheese as the bottom layer (since it must be touching bread), and the final order as ham > lettuce > tomato > potato salad > cheese. 55: This is a little complicated. Judging from each of them, B must be in back left, and A in front left, just because all the others hate the left in some way. With A locked down, C must be in back middle, since A can't be adjacent, and C can't be on the sides. By process of elimination for F, that puts them in back right. That leaves E in front right, and D in front center, for: ADE BCF
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Aug 26, 2019 12:59
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#53 We need to both fill in the 'hole' and connect with the odd L-shape on the right side. To do that, the hole can be turned into a piece that hooks over the top of a three-wide piece, and the rest fills in below that. The right side is cut into a piece that it is three-wide, with a two-wide part below it - it's the L-shape with the top of it extended two squares leftward. #134 The one that is noticeable is D - it's clearly not on a hangar, since that shape cannot be used to hang clothes. So there's probably something more interesting under it, which the professor apparently likes. #054 Ham Lettuce Tomato Potato Salad Cheese #55 ADE BCF This works, it's probably the only solution but I didn't verify that.
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Aug 27, 2019 07:08
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mateo360 posted:wait... is that actually Stachenscarfen's hair? I always though it was suppose to be something like a fur cap. Yeah, that's hair. It's a lot clearer in official art, though, to be fair. Also, update, naturally: 
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Aug 28, 2019 01:08
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137: If you go with the lazy method, 7, as the spot blocks two pieces. However, 8 is the answer - 4 in the same style, and then 4 horizontal pieces. That way, you only lose 1 piece to the hole. 56: I think the only answer is 3+3+3 = 9 (upside-down 6)? 57: Answer should be 6 and 9. Unlike the others, no amount of rotating can save these. 0, 1, and 8 are fine as is. 2 and 5 can be rotated 180 degrees and swapped. 3 can be rotated 180 degrees. 4 can be rotated 90 degrees counterclockwise, and 7 can be rotated 90 degrees clockwise.
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Aug 28, 2019 01:42
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Puzzle 137: 8 more, for 9 pieces total. Starting with the untouched bar, split the chocolate into two 5x5 squares, and divide the one with no hole into 5 pieces vertically. Divide the one with the hole into 5 pieces horizontally. One piece will have the entirety of the hole, so throw it out. Voila! Puzzle 056: 2+3+4 = 9. You have to turn the 5 and 6 stamps upside-down. Puzzle 057: Send 6 and 9 back. 0, 1, and 8 are good as they are. Rotate 2, 3, and 5 180º. Rotate 4 90º. Rotate 7 at an odd angle.
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Aug 28, 2019 01:52
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Dragonatrix posted:
No, not at all. There are plenty of reasonable, logical and plausible reasons for Future Flora to not have been seen. Ran away when Layton went evil, died, held hostage by Layton/ Willingly going along with him, moved away before he went evil, etc. Whatever Future Flora is doing is of no consequence currently to the bigger problem of evil Layton.
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Aug 28, 2019 04:03
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Hmm... Well, yes, I suppose so... 
But you haven't seen a future Flora here?
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mateo360 posted:No, not at all. There are plenty of reasonable, logical and plausible reasons for Future Flora to not have been seen. Ran away when Layton went evil, died, held hostage by Layton/ Willingly going along with him, moved away before he went evil, etc. Whatever Future Flora is doing is of no consequence currently to the bigger problem of evil Layton. I'm pretty sure that if my surrogate caretaker and friend who consistently leave me behind everywhere went into the future, and proceeded to not even think about the fact that my future self is nowhere to be found I'd be pretty miffed too. It's not really relevant to the bigger problem, but it's a pretty hurtful thing anyway, especially considering their track record with actually paying any attention to Flroa at all.
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Aug 28, 2019 04:38
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137. By cutting four pieces on the left side horizontally, you can cut 8 more pieces instead of just 7. 56. 3+3+3=9 57. Only plates 6 and 9 need to be sent back. 0, 1, and 8 are perfectly rotationally symmetrical, 2 and 5 turn into each other when rotated 180 degrees, and 4 rights itself if rotated 90 degrees counterclockwise. 7 is probably readable enough rotated some amount or another.
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Aug 28, 2019 06:14
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I'm still left wondering what that thing item of clothing D is attached to is. Didn't really check before, but it seems so far Layton hasn't made the claim to actually be in the future (he's just acknowledging that other people say it). #137 8 more. Cut four pieces horizontally from the side with the hole in it. Then four more vertically from the rest #056 3 + 3 + 3 = 9, or 1 + 1 +1 = 3 if you use the '7' stamp right. #057 Only 6 and 9. 2 & 5 can be swapped and rotated, the rest can be rotated (0,1,8: 360 degrees, 3: 180, 4:90 left, 7: 135 right ).
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Aug 28, 2019 08:00
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Polderjoch posted:I'm pretty sure that if my surrogate caretaker and friend who consistently leave me behind everywhere went into the future, and proceeded to not even think about the fact that my future self is nowhere to be found I'd be pretty miffed too. Flora deserves better. 
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Aug 29, 2019 03:37
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137: Do two horizontal rectangles above the hole, and two below, so you get four pieces out of the left 5 cm of the bar. Then do four more vertical breaks in the right-hand 4 cm of the bar. Total: 8 pieces that don't have any part of the hole in them. 056: The stamps are for digits 3, 4, 5, 6, and 7, but if you realize that the 6 can be turned around to make 9, you've already solved half the puzzle. 3+3+3=9 using just two stamps. 057: The 0 and 1 are symmetrical; the 2 can be rotated 180° to become a 5, and the 5 can be rotated to become a 2. The 3 can rotate 180° to look right. The 4 is unusable. And so are the 6 and 9: you'd think they would be useable for each other like in the last puzzle, but since they are rotational images of each other, if they are flipped then they will be rotational images of each other's flipped version. The 7 can be used by rotating it until the other leg is horizontal, at which point it will look right. And the 8 is symmetrical. So the unfixable digits are 4, 6, and 9: 3 digits in total. (Update: on looking at other people's answers I see that I missed that the 4 can be rotated 90° to fix it, so the correct answer was that just the 6 and 9 are unusable).
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Aug 29, 2019 15:10
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I dunno about anyone else, but to me it sure feels like we've been waiting a lot longer for this than it actually has been.
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Aug 30, 2019 00:49
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Ooh, things are getting spicy! 059: Red =>, Green <<=, Blue <=, Red =>> (dest), Yellow =>>, Green <= (dest), Blue <<= (dest), Yellow => (dest). 060: A.  062: Hoooo. White <<=, Yellow <=, Red =>> (dest), White =>, Black =>>, Green <=, White <<=, Yellow <=, Black =>> (dest), Yellow <=, White =>> (dest), Green =>, Yellow <<=, Green =>, Blue =>> (dest), Yellow <= (dest), Green <<= (dest), and I get the sneaking feeling it's possible to do better. That's 17 moves, if you were counting.
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Aug 30, 2019 01:50
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quote:Possibly, but maybe later. This seems to be a misplaced image. 59: Green left for YRG-B. Red right over green for Y-GRB. Yellow right for -YGRB. Green left over yellow to GY-RB. Blue over red to GYBR-. Red right to GYB-R. Yellow over blue to G-BYR. Blue left to finish at GB-YR. I think you can also do it starting with Red left as well. 60:  So it's A. Starting with C, then moving on to B makes it pretty clear. 62: This is honestly not too hard, but it's just long. I think the optimal way of doing it was already listed by Quackles, given it's designed so you need to get the yellow frog on the right all the way to the left. Thus, trying to vary it too much (in the opposite direction, for example) would waste moves. I came up with an alternate method, but it also involves 17 moves. (Just listing the colours of the moved frogs, it's White, Orange, Red, White, Orange, White, Green, Orange, Black, Blue, Orange, Green, White, Black, Blue, Green, Blue.
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Aug 30, 2019 03:03
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#059 (numbered left to right, * means ribbon spot) in 8 moves: 4-3, 2-4, 1-2, 3-1*, 5-3, 4-5*, 2-4*, 3-2* #060 I'm assuming the plots are contiguous, although it's not specified. The well condition means each must have one of the four squares around the well, one of which is the star. C must have the north one, so it's out. B must take the west one; if A or D takes it then B's plot must surround C, and C has only one square. A cannot take the east square since it's more than six squares away from the house, so the answer must be A. #062 My best is 15 moves: 6-4,5-6,7-5,6-7*,4-6,2-4,3-2(*),5-3,6-5*,4-6*,2-4,3-2,1-3*,2-1*,4-2* Kangra fucked around with this message at 04:36 on Aug 31, 2019 |
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Aug 30, 2019 06:46
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Cyouni posted:This seems to be a misplaced image. Yeah, it should've been  Fixed now though, thanks.
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Aug 30, 2019 07:29
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059: Here's an efficient order to do things in: - Red hops from 2 to 3 - Green jumps over red, from 4 to 2 - Blue hops from 5 to 4 - Red jumps over blue, from 3 to 5 - Yellow jumps over green, from 1 to 3 - Green hops from 2 to 1 - Blue jumps over yellow, from 4 to 2 - Yellow hops from 3 to 4 060: Without doing any drawing, I can tell that the answer is going to be A, because all four squares touching the well must belong to different houses. The square N of the well belongs to C. If B doesn't have the square W of the well, there's no way to extend his portion to the south, which means that to touch the well his portion would have to stretch over to the E side, wrapping around C's house in a way that would make it impossible to assign C a contiguous portion. And A is at least seven squares away from the E side of the well (counting his own house), so the east side of the well can only belong to D. By elimination, the S side of the well where the star is will belong to A. And knowing that, it became easy to come up with the only shape that could possibly work:  062: I came up with a seventeen-move solution; don't know if it's the most efficient one, but the puzzle didn't actually say "Find the most efficient solution", so here it is: White jumps 6-4 Yellow hops 7-6 Red jumps 5-7 White hops 4-5 Black jumps 2-4 Green hops 3-2 White jumps 5-3 Black hops 4-5 Yellow jumps 6-4 Black hops 5-6 White jumps 3-5 Yellow hops 4-3 Green jumps 2-4 Yellow hops 3-2 Blue jumps 1-3 Yellow hops 2-1 Green jumps 4-2
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Aug 30, 2019 09:12
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The final image before the end-of-update puzzles also seems to be a repeat.
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Aug 30, 2019 11:40
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The only one I actually solved yet is 59: YR_GB (Start) Y_RGB (1) YGR_B (2) YGRB_ (3) YG_BR (4) _GYBR (5) G_YBR (6) GBY_R (7) GB_YR (8)
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Aug 31, 2019 07:43
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 Chinatown is a pretty lengthy stretch here, really. This is the shorter bit, but it does have so, so many puzzles...
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Sep 2, 2019 00:46
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063: Weird order aside, he's still counting in groups of five, so you just have to match the number in the ones place to the correct finger. In the case of 763, it's the pinky. 064: C said "D is lying" before D even said anything. Very fishy, so C is the culprit. 065: Each shape is composed of numbers—3, 4, 5, 6, and 7. Every shape except for the bottom-left one is made using its number a number of times equal to its value (three 3s, four 4s, etc.) so the bottom-left one is the odd one out. 066: Vase A matches the man's profile (as in that very famous optical illusion) when flipped upside-down. 067: One move. Move the leftmost bowl into the empty space on the right. 068: Unfolding paper A creates a diamond with sides that are two squares wide, so paper B must be cut in a straight line from the leftmost dot in the middle row to the lowest dot in the middle column. 069: Ruby is on the end and Sapphire is next to Ruby, so Sapphire must be B or D. Ebony is three over from Sapphire, so Ebony must be on the other end. Jade is not green, so Jade must also be B or D, which leaves Pearl as C. 070: The third child from the bottom is the eighth child aka Luke, so the third child from the bottom is a boy. 074: Left to the wall -> two spaces right -> down to the wall -> up two spaces -> right to the wall -> up two spaces -> left two spaces -> up one space -> left to the wall -> down one space -> left to the wall -> up one space -> left out the exit. 148: Zero matches. Turn the equation upside-down and it becomes 10 = 9 + 1.
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Sep 2, 2019 01:46
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#064 C is pretty quick to cast blame on someone else, before D has even said anything. Obviously a time traveler, and they cannot be trusted. #063 The 3 finger (i.e. pinky) #074 Basically amounts to 'get to the second street from the left and clear bananas on it until you can walk to the exit'. #148 0. Turn it upside down. #065 The one that's made from 5 6's (lower left), instead of the others which have the number of the numeral they're made from. #066 A, because the man is walking upside down. #067 1. Move the leftmost bowl to the right end. #068 Using (0,0) as the lower left, cut a diagonal line from (0,2) to (2,0). #069 Using #5 & #1, we narrow the choices for Ruby to (AE) and Sapphire to (BD). #2 yields Ebony can be (ABCD) and #3 gives Jade as (ABDE). #3 means Ebony can only be A or E, but since it can't be E, Ebony must be A. Then Sapphire is D, Ruby is E, and Jade is B (since all other choices are taken). Therefore Pearl is C. Nice little puzzle. e:Forgot the last one: #070 Whatever Luke identifies as, since he's the third from the bottom. Moreover, If we assume Luke is a son, we can specify all the children from the conditions: Dn != 8 (since Luke is there) Dn > 2 (all daughters must be younger than the two eldest sons) S3 > 3 => S1 = 1, S2 = 2, D1 = 3 S4 < D2 (and D1 < S4, but we already knew that) => S3 = 4, S4 = 5 D4 = D3 + 1 => D2 = 6, S5 = 7, Luke = S6, D3 = 9, D4 = 10 Kangra fucked around with this message at 03:42 on Sep 2, 2019 |
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Sep 2, 2019 03:23
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That's a few puzzles. 64: So this is a spot-the-liar one. A can't be the liar, because that'd make B lying. Similarly, B can't be lying, because then A would be the culprit. The other two are the opposite, where either invalidates the other. However, since it's in order...C says D's lying before D even says anything, so it has to be C who's the liar. 63: Since it's a repeating pattern, it's the 3 finger, or the pinky. 74: Go left until the banana and slide to the end. East three steps, then slide south. After that, just avoid all the bananas and walk to the exit. 148: Changing XI to IX, or making the equation 1+9=10, is enough, so one matchstick. 65: The bottom left one is the only one that's not constructed out of equivalent numbers. It's a shape made out of five 6s. 66: A, once you flip the vase and look at the negative space. 67: One move, moving the first one to the other end. 68: Middle dot on left to middle dot on bottom. 69: Green is not green would be a bit weird to read. First one combined with the last two clues means it has to be Ruby/Sapphire/other two/Ebony, or the reverse. However, Ebony is not white, so it's Ebony/other two/Sapphire/Ruby. Then Jade is not green, so the green cat must be Pearl, or C. 70: Luke, Luke, the eighth child is the third from the bottom. And you're male, right?
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Sep 2, 2019 06:33
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064: This sounds like a classic truth and lies logic puzzle, but if you approach it as such, it's unsolvable. The key is to notice the words "in order" in the puzzle setup. C said that D was lying before D had spoken yet. That's suspicious, and so the puzzle setup tells us to blame C. 063: This is modulo-5 math, so the answer is finger 3, the pinky. 074: Walk forward (west), slide until you bump. Take two steps back east, then go south and slide. Now there's a peel-free route to the blue arrow. 148: Move one matchstick from the XI to the X, so that it becomes I + X = XI. No special tricks needed for this one. 065: The bottom left is out of place. All the others are made of numbers, repeated that number of times (e.g. the top left shape is a 3 repeated 3 times). But the bottom left shape, whether it's a 6 or a 9, is only repeated 5 times. 066: Vase B has a shape that resembles the man: a big belly and a plump "head" perched on it, with no neck to speak of. 067: Pick up the leftmost bowl and put it to the right of the rightmost. One move. 068: I don't feel like drawing so I'll describe. If the dots are a 5x5 grid with A-E columns and 1-5 rows (from left and bottom respectively) then you'd cut along the lines B3-C4-D3-C2-B3. Update: This is wrong. That's what it would look like if you were cutting the piece out of an entire square. I failed to remember that the second piece is folder as well, so the other answers given (which I looked at after giving mine) are correct here. 069: Ruby is on the end, is next to Sapphire, and Sapphire is three over from Ebony. That means that Ebony is on the opposite end from Ruby. Since Ebony is not white, Ruby isn't the red one so Ruby is E, Sapphire is D, and Ebony is A. Since Jade isn't green, Jade is B and that means that Pearl is C by elimination. 070: Trick question. The puzzle setup said that Luke is the 8th child, which in a family of ten is the third child from the bottom! Luke is a boy, of course. Tax Refund fucked around with this message at 14:49 on Sep 2, 2019 |
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Sep 2, 2019 14:46
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Ooh, that's a lotta puzzles. 064: Wait, C said D is lying before he even heard what she had to say? That can't be right. I say C broke the window. 063: His pinky. 'cause it's tagged '3'. 074:  148: Zero! Just look at it upside down. That makes X = IX + I (10 = 9 + 1). 065: Bottom left. Each of the others is a number repeated in rotational symmetry that many times (so the '3' is repeated 3 times around a circle, the '7' is repeated 7 times, and so on). But the bottom left is a '9' repeated only 5 times. 066: B... I think. They're the same shape! 067: 1 move. Move the bowl of plain noodles on the left end to the right end. What? No one said you could only move to adjacent locations! 068: Please don't ask me to explain this one.  069: OK, so first off, Ruby is on either end, and Sapphire is next to her. So Ruby is at either the left or right end, and Sapphire is in the 2nd or 4th place. Sapphire is three over from Ebony, and since Sapphire is in the 2nd or 4th place, Ebony must be on the other end. But Ebony is not white, so Ebony must be the red cat on the left end, and Ruby must be the white cat on the right. Sapphire must thus be the black cat in the 4th place. The center cat is not Ruby, Ebony, or Sapphire. It also can't be Jade, because it's green and Jade specifically isn't green. So, the center cat (C) is Pearl. Finally, Jade must be the blue cat in the 2nd place. What confusing names! 070: Oh, very funny. The third child from the bottom is the eighth child from the top. Which is Luke. Who's a boy.
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Sep 3, 2019 00:01
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Probably no update later today, unfortunately. Just because it's bound to end up being pretty long and I haven't had much time to work on it. Ideally be done by Thursday/Friday, though. Not saying it's impossible to be ready for today, but... I'm not counting on it.
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Sep 3, 2019 18:36
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We're rooting for you! Take whatever time you need.
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Sep 3, 2019 21:33
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Took a tiny bit longer than I expected, but it's fine really:
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Sep 9, 2019 10:40
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Ooh! It's a good evening for a puzzle. 071: The can of white paint. 'nuff said. 072: Red. This'd take too long to explain, but let's say it's because circle A is a certain distance from the painted circle. 073: Let’s start by placing the blue tiles. There’s four blue tiles total, so one must be on each row and column. In the top row, the only free space that isn’t next to the existing blue tile is in column 2. From there, we realize that the blue tile in row 3 has to be on the far left, in column 1; that lets us put the last blue tile in row 4, column 3. There’s now only three slots in which we can put green tiles: on the top right corner, and in either of the spaces on the left column. Since we can only have one green tile per column, we know the top right corner space MUST have a green tile. From there, we can narrow the other green tile’s position to row 2 on the far left (because the bottom left corner would be in a diagonal line with the top left). At this point, there are only two places we can put red tiles: the bottom left corner, and the third row on the right. We have only yellow and orange tiles left. The top row already has a yellow tile, so the free space there must be orange. The bottom row already has an orange tile, so the free space there must be yellow. From there, the last two tiles seat themselves easily. Y, B, O, G G, R, Y, B B, O, G, R R, Y, B, O
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Sep 9, 2019 11:20
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071: The white paint, of course. 072: No need to paint the entire pyramid. The adjacent two to the lower-right of the red one are blue and yellow -> the piece of wood that touches both of them must be red -> the two next to A both touch that red one, so they must be blue and yellow -> A must be red. 073: Bit tricky to do without visually putting them in place as you go, but I'm gonna give it a shot anyway! Four blue tiles, so let's do blue first: only possibility for the top row is column two, which makes the only possibility for the third row column one, which puts the fourth row's blue tile in column three. Red must be in R1C4, R3C4, or R4C1. There can only be one in the fourth column, so one of them has to be in R4C1. That leaves only two slots left for green, so that's where they are: R1C4 and R2C1. The last red one must be in R3C4, then, which leaves the yellows in R2C3 and R4C2 and the oranges in R1C3 and R3C2. tl;dr (from left to right) Row 1: yellow, blue, orange, green; Row 2: green, red, yellow, blue; Row 3: blue, orange, green, red; Row 4: red, yellow, blue, orange.
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Sep 10, 2019 01:42
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#071: The White bucket. #072: Red. Since any small triangle of three necessarily has all three colors in it, then a parallelogram of 4 has to have the same color on the 'far ends'. Which means the log in a line between A and the red log must also be red, and applying that logic again means A must be red. #073: All colors need to be used, and in particular that means every column and row must have a blue somewhere in it, since there are four blues. The primary way to proceed is to find tiles that have four colors in the same row, column, or diagonal, as the fifth color must go in that tile. Numbering using (row, column) from the upper left (1,1), then the top of the third column (1,3) cannot be yellow, red, blue, or green, and must be orange. That means (2,3) is surrounded by four colors and must be yellow. The last tile in column 3 must be blue. Continuing to (3,4), it has blue, green, and orange, with yellow adjacent diagonally, so must be red. That leaves green as the only color for (1,4). In column 2, the lower two tiles cannot be blue since (4, 3) is. That means (1,2) is blue. (3,2) is surrounded by red and green on the sides, and blue and yellow diagonally, so must be orange. This leaves only yellow for (4,2), since there is a red in the column above it, and green, orange, and blue adjacent. Finally, row 3 in column 1 must have the last blue tile. (2,1) has yellow and red adjacent, and blue and orange diagonally adjacent, and must be green. The last tile (4,1) has yellow, blue, and green above it, and with orange at both (4,4) and (3,2) it has to be red. Final layout: YBOG GRYB BOGR RYBO e:Corrected, see below (it could only be on a diagonal). Kangra fucked around with this message at 19:24 on Sep 10, 2019 |
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Sep 10, 2019 16:00
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Kangra posted:
Objection, there was zero chance - Because there are four blue tiles and three of all the others, a blue tile must be in every row, column, and diagonal; you mentioned this yourself. Therefore it's impossible for ORGY to turn up in any row, column, or diagonal of a correct answer.
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Sep 10, 2019 18:34
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071: Black: 7+8+6 = 23 bricks. Red: 1+5+7+8 = 21 bricks. Yellow: alignment isn't perfect, but call it a square of 2.5 bricks × 5 rows, plus half of a square of 2 (or 2.5) bricks by 5 rows. So that's 12.5 + 6.25 at most, or 18.75 at most. Green: 4+12+4 = 20 bricks. Blue: 4.5 x 3 in lower rectangle, then the two triangles add up to exactly half of a 4.5 × 4 rectangle (which would be 4.5 × 2), so total is 4.5 × 6 = 27 bricks. The blue paint was used the most. 072: I can tell it's going to be red without doing the paint work. The six circles around the red one are going to be alternating blue and yellow, so the one a "hexagonal knight's move" right and down is going to be touching a yellow and a blue circle. It will therefore be red, it will be surrounded by blue and yellow, and the circle that's a hexagonal knight's move right and down from it will also be red. Since that's the one marked A, there's our answer: red. 073: I'll call the columns A-E and the rows 1-4, so A4 is yellow, B3 is red, etc. Well, B1 is aligned with four colors already, so it has to be yellow. And C4 is also aligned with four colors, so it has to be orange. Now C3 obviously must be yellow. That then puts D2 aligned with four different colors so it must be red. D4 now must be green, then B4 must be blue, as must C1. Now B2 must be orange. In the A column, A1 must be red, A3 must be green, and A2 must be blue. Which leaves a surprisingly pretty arrangement.
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Sep 13, 2019 17:30
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| # ? May 3, 2024 22:50 |
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Maybe I’m missing something, but can’t you solve #72 without any logic whatsoever? The only colour used in the puzzle is red, so the only guaranteed solution can be red, since otherwise it could either yellow or blue depending on which colour you opt to use?
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Sep 13, 2019 23:19
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